Question

$\underset{x\to \sqrt{10}}{\lim }\frac{\sqrt{7+2x}-\left(\sqrt{5}+\sqrt{2}\right)}{x^2-10}$

Answer 1

$\underset{x\to \sqrt{10}}{\lim }\left[\frac{\sqrt{7+2x}-\left(\sqrt{5}+\sqrt{2}\right)}{x^2-{\left(\sqrt{10}\right)}^2}\right]$

= $\underset{x\to \sqrt{10}}{\lim }\left[\frac{\sqrt{7+2x}-\sqrt{{\left(\sqrt{5}+\sqrt{2}\right)}^2}}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)}\right]$

= $\underset{x\to \sqrt{10}}{\lim }\left[\frac{\sqrt{7+2x}-\sqrt{5+2+2\sqrt{5}\sqrt{2}}}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)}\right]$

= $\underset{x\to \sqrt{10}}{\lim }\left[\frac{\sqrt{7+2x}-\sqrt{7+2\sqrt{10}}}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)}\right]$

Rationalising the numerator:

$\underset{x\to \sqrt{10}}{\lim }\left[\frac{\left(\sqrt{7+2x}-\sqrt{7+2\sqrt{10}}\right)\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}\right]$

= $\underset{x\to \sqrt{10}}{\lim }\left[\frac{\left(7+2x\right)-\left(7+2\sqrt{10}\right)}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}\right]$

= $\underset{x\to \sqrt{10}}{\lim }\left[\frac{2\left(x-\sqrt{10}\right)}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}\right]$

= $\frac{2}{\left(\sqrt{10}+\sqrt{10}\right)\left(2\sqrt{7+2\sqrt{10}}\right)}$


$$\begin{gathered} =\frac{2}{\left(2\sqrt{10}\right)\times 2\sqrt{7+2\sqrt{10}}} \\ =\frac{1}{2\sqrt{10}\sqrt{{\left(\sqrt{5}+\sqrt{2}\right)}^2}} \end{gathered}$$

= $\frac{1}{2\sqrt{10}\left(\sqrt{5}+\sqrt{2}\right)}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{1}{2\sqrt{10}}\left[\frac{\sqrt{5}-\sqrt{2}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2}\right]$

= $\frac{\sqrt{5}-\sqrt{2}}{6\sqrt{10}}$